Question: $f(x) = \begin{cases} 4x^3+2 & \text{for} ~~~~x\gt{-1} \\ 3x+1& \text{for} ~~~~ x \leq-1\end{cases}$ Evaluate the definite integral. $\int^0_{-4}f(x)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $-19$ (Choice B) B $-\dfrac{37}{2}$ (Choice C) C $20$ (Choice D) D $\dfrac{41}{2}$
Splitting up the definite integral Since we're working with a piecewise function, we need to split the definite integral up into two pieces: $\phantom{=} \int^0_{-4}f(x)\,dx$ $= \int^0_{-1}f(x)\,dx + \int^{-1}_{-4}f(x)\,dx~~~~~~$ [Why did we split at -1?] $= \int^0_{-1}(4x^3+2)\,dx + \int^{-1}_{-4}(3x+1)\,dx ~~~~~~$ Evaluating each piece Next, let's evaluate each of these definite integrals one at a time. The first definite integral: $\begin{aligned} \int^0_{-1}(4x^3+2)\,dx &=(x^4+2x)\Bigg|^0_{{-1}} \\\\ &= \left[ ( 0)^4+2\cdot(0) \right] - \left[ ( {-1})^4+2\cdot({-1}) \right] \\\\ &= \left[0\right] -\left[-1 \right] \\\\ &= {1}\end{aligned}$ The second definite integral: $\begin{aligned} \int^{-1}_{-4}(3x+1)\,dx &=\left(\dfrac32x^2+x\right)\Bigg|^{-1}_{{-4}} \\\\ &= \left[\dfrac32 ( {-1})^2 +( {-1}) \right] - \left[\dfrac32 ( {-4})^2 +( {-4})\right] \\\\ &= \left[\dfrac12\right] -\left[20 \right] \\\\ &= {-\dfrac{39}{2}}\end{aligned}$ Putting the pieces together Now let's add these two pieces together to find the answer: $\phantom{=} \int^0_{-1}(4x^3+2)\,dx + \int^{-1}_{-4}(3x+1)\,dx$ $ = {1} + \left({-\dfrac{39}{2}}\right)$ $ = -\dfrac{37}{2}$ The answer $\int^0_{-4}f(x)\,dx = -\dfrac{37}{2}$